Homework Help:
Thermodynamic prob

A heat Engine recieves 6 kg/h gasoline which has a calorific value of 42000 kj/kg. If heat rejected from the engine is 40 kw, calculate
a- the power done by the engine
b- the engine effiecincy

-this wt i got today in my exam and unfortuntely i didnt study it well but i did the following :
1- 6 kg/h -----> 1/600 kg/s
then i multiplied it by the 42000 kj/kg
so i can get kj/s ( which is the power )
so is that correct ??

2- i said effieciency = w/Qh = 1 - Qh/Ql

and Qh = W - Ql = 42000 - 40

so is that too correct ?

* i knw i didnt study well and sorry if i posted it in a wrong section